Some comments from my side:
1.) The stability check in the BODE diagram concerns the LOOP GAIN response only (because once you did mention "closed-loop system" in your text.)
2.) The shown system is "conditionally stable". That means: It is stable - regardless the properties at the frequency A. However, if you REDUCE the gain within the loop until the gain crosses the point A (the phase remains unchanged) the closed-loop system will be unstable.
Such conditional stable system should be avoided because a gain reduction can happen due to aging or other damping effects. Remember: Classical feedback systems with a continuos decreasing loop phase will become unstable (under closed-loop conditions) for rising gain values (beyond a certain limit) only.
As to your next question - the input signal Vi does not influence stability properties at all. Stability is determined by the loop components only. That is the reason, we investigate the loop gain only.
EDIT: Here is an explanation why the closed loop (your example) will be stable:
If a closed-loop system is unstable, this point of instability also must be "stable". That means - either we will have "stable" and continuous oscillations or the output is latched at one of the supply voltage rails. In both cases, this point of instability is fixed.
Now - what happens at the point A in your example? Here we have a rising phase which is identical to a NEGATIVE group delay at this point (group delay is defined as the negative phase slope). This is an indication for the unability of the closed-loop system to let the amplitudes rise (oscillations or latching at the supply rail). Rather, the system returns to a stable operating point.
A final information: The stability check investigates either (a) the -180deg line or (b) the -360 deg line. This depends on what you are investigating: (a) Either the simple product GH or (b) the loop gain LG which is LG=-GH.
Using your system as an example, it's interesting to consider what happens during the first 10sec, or so, following the application of a unit step at the system input (this may convince you that it's far easier to base a stability analysis on the open-loop!)
\$\small 0<t<5\$: The output from the delay is zero; the error signal is unity, hence the system output is a unit ramp (integral of step = ramp). Hence, the output reaches 5 at t=5.
\$\small t=5\$: the unit ramp begins to emerge from the delay
\$\small 5<t<6\$: The unit ramp subtracts from the unit step, hence the error signal ramps down from 1 and reaches zero at t=6. As the integrator input is now ramping downwards, the integrator output is no longer a ramp, it's a parabola, gradually diverging from the original ramp (integral of a ramp is a parabola). The integrator output reaches 5.5 at t=6.
\$\small 6<t<10\$: The error signal is now negative, and is still a negative-going ramp with a gradient of -1. The integrator output decreases parabolically from 5.5, reaching -2.5 at t=10.
\$\small t>10\$: The parabolic sections of the integrator output now begin to emerge from the delay and subtract from the unit step input. Note that, when the delay output signal goes negative, the error signal will be >1 and the integrator output signal will exceed the earlier ramp in magnitude. The system is unstable.
Choosing a smaller delay time (or applying a fractional integrator gain) will render the system stable (try it, if you've got the odd day to spare!)
In contrast, if the delay is replaced by a 1st order lag, the feedback path is far less aggressive. The output from the lag starts to grow exponentially from t=0 and begins to reduce the error signal immediately. This means that the error signal falls exponentially, from t=0, and the system output grows in a much more leisurely fashion. The worst case is where the lag is replaced by an integrator, giving rise to a 2nd order system with zero damping and an oscillatory response. This is critical stability, and things can't get any worse from a stability perspective. Therefore the 2nd order system with poles in the LH s-plane cannot be unstable. In terms of electrical components, it's an LC circuit without any R.
Best Answer
"When an element is unstable , it doesn't mean it is unstable for all inputs. For certain inputs the block can still produce stable outputs."
At first, an "element" (a part) cannot be unstable. It is better to use the term "active block" or "amplifier". Hence, an amplifier can be unstable if it has feedback, which produces a closed-loop pole (pair) with a positive real part. This is true for all inputs because it is the feedback loop which produces self-excitement of the system - not the input signal.
"During feedback, the input given to the the block gets modified such that output doesn't blow up."
An unstable system can be stabilized if the overall negative feedback overrides (dominates) the positive feedback which would cause instability (without negative feedback). In your case (example), the gain block has an inherent positive feedback (causing a pole at +0.5). This gain block ist stabilized using - as shown - 100% negative feedback causing a total closed-loop gain of "2 (and a pole at "-0.5")